On the rigidity of certain Pham-Brieskorn rings

TL;DR

This paper investigates the rigidity of Pham-Brieskorn rings, a class of algebraic structures, providing partial results towards a conjecture about their properties when certain conditions on exponents are met.

Contribution

It offers new partial results and insights into the conjecture regarding the rigidity of Pham-Brieskorn rings under specific conditions.

Findings

Partial confirmation of the rigidity conjecture for certain cases.

Identification of conditions under which Pham-Brieskorn rings are rigid.

Advancement in understanding the structure of these algebraic rings.

Abstract

Fix a field $k$ of characteristic zero. If $a_1, ..., a_n$ ($n>2$) are positive integers, the integral domain $B = k[X_1, ..., X_n] / ( X_1^{a_1} + ... + X_n^{a_n} )$ is called a Pham-Brieskorn ring. It is conjectured that if $a_i > 1$ for all $i$ and $a_i=2$ for at most one $i$, then $B$ is rigid. (A ring $B$ is said to be rigid if the only locally nilpotent derivation $D: B \to B$ is the zero derivation.) We give partial results towards the conjecture.

Equations

$$T_{n}=\big{\{}\,(a_{1},\dots,a_{n})\in\mathbb{Z}^{n}\,\mid\,\text{$a_{i}\geq 2$ for all $i$ and $a_{i}=2$ for at most one $i$}\,\big{\}}.$$

$$R_{0}=\bigcup_{i\in G}\big{\{}\,\textstyle\frac{b}{s}\,\mid\,\text{$b\in B_{i}$ and $s\in B_{i}\cap S$}\,\big{\}}.$$

$$J^{\ast }(S)$$

$$J(S)$$

$$\mathrm{type}(S)=|J^{\ast }(S)|\text{and}\mathrm{cotype}(S)=|J(S)|.$$

$$B_S=B_{a_1,\dots ,a_n}=\mathbf{k}[X_1,\dots ,X_n]/⟨X_1^{a_1}+\cdots +X_n^{a_n}⟩.$$

$$\gcd (d_1,\dots \widehat{d_i}\dots ,d_n)\ne 1⟺\mathrm{lcm}(a_1,\dots \widehat{a_i}\dots ,a_n)\ne \mathrm{lcm}(S)⟺i\in J(S).$$

$$A=R[u]/⟨f+u^g⟩\text{ and }B=R[z]/⟨f+z^n⟩$$

$$B$$

$$S_i=S_i^{\prime }\text{and}{g}_i(S^{\prime })\mid a_i\mid a_i^{\prime }.$$

$$S_n=S_n^{\ast }=S_n^{\prime },g\mid a_n^{\ast }\mid a_n\text{and}g\mid a_n^{\ast }\mid a_n^{\prime },$$

$$A=R[u]/⟨f+u^{a_n^{\ast }}⟩,B=R[z]/⟨f+z^{a_n}⟩,B^{\prime }=R[w]/⟨f+w^{a_n^{\prime }}⟩.$$

$$J(S)=\big{\{}\,j\,\mid\,\text{$S$ is not minimal with respect to $<^{j}$}\,\big{\}}.$$

$$(a_1,\dots ,a_{j-1},g_j(S),a_{j+1},\dots ,a_n){<}^{j}S.$$

$$I(S)=\{1,\dots,n\}\setminus J(S)=\big{\{}\,i\,\mid\,\text{$a_{i}$ divides $\operatorname{{\rm lcm}}(S_{i})$}\,\big{\}}.$$

$$S^{\nu }=\text{componentwise product }(1,\dots ,1,p_{j_{\nu }}^{e_{j_{\nu }}},1,\dots ,1)\cdot S^{\nu -1}$$

$$\ast$$

$$\Delta_{M}(S)=\frac{\operatorname{{\rm lcm}}(S)}{\gcd\big{\{}\,\operatorname{{\rm lcm}}(S_{j})\,\mid\,j\in M\,\big{\}}}\qquad\text{if $M\neq\varnothing$}.$$

$${\Delta }_M(S)={\Delta }_{J(S)\cap M}(S)\text{and}{\Delta }_M(S)=1\iff M\cap J(S)=\varnothing ,$$

$$D\in \mathrm{HLND}(B_S)\setminus \{0\}$$

$$\delta >1$$

$$(d_1^{\prime },\dots ,d_n^{\prime })=(d_1,k{d}_2,\dots ,k{d}_n)\text{and}\gcd (d_1,k)=1.$$

$$\phi :B_S=\mathbf{k}[X_1,\dots ,X_n]/⟨X_1^{a_1}+\cdots +X_n^{a_n}⟩\to B_{S^{\prime }}=\mathbf{k}[Y_1,\dots ,Y_n]/⟨Y_1^{a_1^{\prime }}+\cdots +Y_n^{a_n^{\prime }}⟩.$$

$$L$$

Full text

Research of both authors supported by grant 04539/RGPIN/2015 from NSERC Canada.

1. Introduction

If $B$ is a commutative ring of characteristic zero, a derivation $D:B\to B$ is locally nilpotent if for each $x\in B$ there exists $n>0$ such that $D^{n}(x)=0$. If the only locally nilpotent derivation $D:B\to B$ is the zero derivation, one says that $B$ is rigid. One says that $B$ is stably rigid if for any $N\geq 0$ and any locally nilpotent derivation $D:B[X_{1},\dots,X_{N}]\to B[X_{1},\dots,X_{N}]$ (where $B[X_{1},\dots,X_{N}]$ is the polynomial ring in $N$ variables over $B$), we have $D(B)=\{0\}$. Note that stable rigidity implies rigidity.

Fix a field ${\rm\bf k}$ of characteristic zero. If $a_{1},\dots,a_{n}$ ($n\geq 3$) are positive integers, the integral domain $B_{a_{1},\dots,a_{n}}={\rm\bf k}[X_{1},\dots,X_{n}]/\langle X_{1}^{a_{1}}+\cdots+X_{n}^{a_{n}}\rangle$ is called a Pham-Brieskorn ring. These rings and the corresponding varieties have been studied extensively and from several angles; paper [10] refers to [13] for a survey. It is interesting to ask which Pham-Brieskorn rings are rigid or stably rigid. Consider the set

[TABLE]

It is known (and easy to see) that if $B_{a_{1},\dots,a_{n}}$ is rigid then $(a_{1},\dots,a_{n})\in T_{n}$; so one wants to know if the converse is true. The case $n=3$ is settled by:

1.1 Theorem**.**

Let $a,b,c$ be positive integers.

1. (a)

If $(a,b,c)\in T_{3}$ then $B_{a,b,c}$ is rigid. 2. (b)

If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq 1$ then $B_{a,b,c}$ is stably rigid.

This is [10, Theorem 7.1] (the case ${\rm\bf k}=\mathbb{C}$, with part (b) implicit, is [12, Lemma 4]).

For arbitrary $n\geq 3$, one has:

1.2 Theorem**.**

If $\sum\limits_{i=1}^{n}\frac{1}{a_{i}}\leq\frac{1}{n-2}$ then $B_{a_{1},\dots,a_{n}}$ is rigid.

The case ${\rm\bf k}=\mathbb{C}$ of Theorem 1.2 is [9, Example 2.6]. The general case follows, because (as one can see) rigidity over $\mathbb{C}$ implies rigidity over any field of characteristic zero.

In the case $n=4$, the following is known:

1.3 Theorem**.**

Assume that $(a,b,c,d)\in T_{4}$. Then $B_{a,b,c,d}$ is rigid in each of the following cases:

1. (a)

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\leq\frac{1}{2}$** 2. (b)

$\gcd(abc,d)=1$** 3. (c)

$a=b=c=3$** 4. (d)

$a=2$, $b,c,d\geq 3$, $b$ is even, $\gcd(b,c)\geq 3$ and $\gcd(d,\operatorname{{\rm lcm}}(b,c))=2$ 5. (e)

$\operatorname{{\rm cotype}}(a,b,c,d)\geq 2$.

Part (e) of Theorem 1.3 is Theorem 7.2(b) of [5], reformulated in terms of cotype (see Definition 3.9 for the notion of cotype). Part (a) is the case $n=4$ of Theorem 1.2. Parts (b–d) are stated in [5, Theorem 7.1] but are proved in [10, Theorem 8.1] and [1, Corollary 1.9].

This article settles many cases not covered by Theorems 1.1–1.3. In order to avoid giving too many definitions in the Introduction, we only present a subset of our results as Theorem 1.4. Other significant results appear in Section 4. The Pham-Brieskorn rings $B_{a_{1},\dots,a_{n}}$ that appear in Theorem 1.4 are defined over an arbitrary field ${\rm\bf k}$ of characteristic zero, and we always assume that $n\geq 3$.

1.4 Theorem**.**

* ***

1. (a)

If $a\geq n\geq 4$ then B_{\mbox{\scriptsize\underbrace{a,\dots,a}_{n}}} is rigid. 2. (b)

If $\sum_{i=1}^{n}\frac{1}{a_{i}}\leq\frac{1}{n-2}$ then $B_{a_{1},\dots,a_{n}}$ is stably rigid. 3. (c)

If $\sum\limits_{i\in I}\frac{1}{a_{i}}<\frac{1}{n-2}$ then $B_{a_{1},\dots,a_{n}}$ is rigid, where I=\big{\{}\,i\,\mid\,\text{a_{i}$divides$\operatorname{{\rm lcm}}(a_{1},\dots\widehat{a_{i}}\dots,a_{n})}\,\big{\}}.** 4. (d)

If $a,b,c,d\geq 1$ satisfy $a\nmid\operatorname{{\rm lcm}}(b,c,d)$ and $\frac{1}{b}+\frac{1}{c}+\frac{1}{d}<\frac{1}{2}$ then $B_{a,b,c,d}$ is rigid. 5. (e)

If $(a_{1},\dots,a_{n})\in T_{n}$ and $\operatorname{{\rm cotype}}(a_{1},\dots,a_{n})\geq n-2$, then $B_{a_{1},\dots,a_{n}}$ is rigid. 6. (f)

If $k_{1},k_{2},k_{3},k_{4}\geq 1$ are pairwise relatively prime and $a\geq 3$ then $B_{ak_{1},ak_{2},ak_{3},ak_{4}}$ is rigid. 7. (g)

If $k_{1},\dots,k_{n}\geq 1$ are pairwise relatively prime and $a\geq n\geq 4$ then $B_{ak_{1},\dots,ak_{n}}$ is rigid. 8. (h)

If $a_{1},\dots,a_{m}\geq 1$ $(m\geq 1)$ satisfy $a_{i}\nmid\operatorname{{\rm lcm}}(3,a_{1},\dots\widehat{a_{i}}\dots,a_{m})$ for all $i\in\{1,\dots,m\}$, then $B_{a_{1},\dots,a_{m},3,3,3}$ is rigid.

Part (a) (of Theorem 1.4) is Corollary 4.6; the fact that $B_{a,\dots,a}$ is rigid when $n\leq a<n(n-2)$ appears to be a new result (if $a\geq n(n-2)$ then $B_{a,\dots,a}$ is rigid by Theorem 1.2). Parts (b) and (c) are Corollary 6.8 and Proposition 4.12, respectively; these two results strengthen Theorem 1.2. Part (d) is the case “$n=4$” of part (c). Part (e) is Corollary 4.16; it generalizes Theorem 1.3(e). Parts (f) and (g) are Corollary 4.10, and (h) is Example 4.17; parts (f–h) are illustrations of stronger results that cannot be stated in the Introduction.

Let us say a few words about Section 3. It is known that if a ${\rm\bf k}$-domain $B$ is not rigid then its field of fractions $\operatorname{{\rm Frac}}(B)$ is ruled over ${\rm\bf k}$, i.e., there exists a field $K$ such that ${\rm\bf k}\subseteq K\subset\operatorname{{\rm Frac}}(B)$ and $\operatorname{{\rm Frac}}(B)=K(t)$ where $t$ is transcendental over $K$. So, one technique for showing that $B$ is rigid is to show that $\operatorname{{\rm Frac}}(B)$ is not ruled over ${\rm\bf k}$. However, that technique is useless when $B$ admits a nontrivial $\mathbb{Z}$-grading, because then $\operatorname{{\rm Frac}}(B)$ is always ruled over ${\rm\bf k}$: we have $\operatorname{{\rm Frac}}(B)=K(t)$ where $K=\operatorname{{\rm HFrac}}(B)$ is the “homogeneous field of fractions” of $B$. Section 3 shows that if $B$ is a graded ${\rm\bf k}$-domain which is not rigid then, under certain additional hypotheses, $\operatorname{{\rm HFrac}}(B)$ itself is ruled over ${\rm\bf k}$. So, to prove that a graded domain $B$ satisfying certain assumptions is rigid, it suffices to show that $\operatorname{{\rm HFrac}}(B)$ is not ruled. It is this technique that allows us to prove part (a) of Theorem 1.4.

2. Preliminaries

Throughout this work, all rings are commutative and have a multiplicative identity $1$. All ring homomorphisms map 1 to 1. If $B$ is a ring then $B^{*}$ denotes its group of units. If $b\in B$ then $\langle b\rangle$ is the ideal of $B$ generated by $b$, and we use the notation $B_{b}=S^{-1}B$ where $S=\{1,b,b^{2},\dots\}$. By a domain, we mean an integral domain. If $B$ is a domain, its fraction field is denoted $\operatorname{{\rm Frac}}(B)$. If $A\subseteq B$ are domains then $\operatorname{{\rm trdeg}}_{A}(B)$ denotes the transcendence degree of $\operatorname{{\rm Frac}}(B)$ over $\operatorname{{\rm Frac}}(A)$. A subring $A$ of a domain $B$ is factorially closed in B if for all $x,y\in B\setminus\{0\}$ we have the implication $xy\in A\Rightarrow x,y\in A$. If ${\rm\bf k}$ is a field, then a ${\rm\bf k}$-domain is a domain that is also a ${\rm\bf k}$-algebra. An affine ${\rm\bf k}$-domain is a ${\rm\bf k}$-domain that is finitely generated as a ${\rm\bf k}$-algebra.

If $A$ is a subring of a ring $B$, we write $B=A^{[n]}$ to indicate that $B$ is isomorphic to the polynomial algebra in $n$ variables over $A$. If $K/{\rm\bf k}$ is a field extension, $K={\rm\bf k}^{(n)}$ means that $K$ is purely transcendental over ${\rm\bf k}$, of transcendence degree $n$.

We write $``\subseteq"$ for inclusion and $``\subset"$ for proper inclusion.

Let $B$ be a ring and $D:B\to B$ a derivation. We say that $D$ is irreducible if the only principal ideal of $B$ containing $D(B)$ is $B$ itself. We say that $D$ is locally nilpotent if for each $b\in B$ there exists an $n\in\mathbb{N}$ such that $D^{n}(b)=0$. A slice of $D$ is an element $t\in B$ such that $D(t)=1$. A preslice of $D$ is an element $t\in B$ such that $D(t)\neq 0$ and $D^{2}(t)=0$.

2.1 Definition**.**

If $B$ is a ring, the set of locally nilpotent derivations $D:B\to B$ is denoted $\operatorname{{\rm LND}}(B)$. If $\operatorname{{\rm LND}}(B)=\{0\}$, we say that $B$ is rigid.

2.2**.**

Let $B$ be an integral domain of characteristic zero, let $D:B\to B$ be a derivation, and let $A=\ker D$. The following facts are well known (refer to [8], [11] or [3]).**

1. (a)

If $D$ is locally nilpotent, then $A$ is a factorially closed subring of $B$. Consequently, if $D$ is locally nilpotent and ${\rm\bf k}$ is a field included in $B$ then $D$ is a ${\rm\bf k}$-derivation. 2. (b)

Assume that $\mathbb{Q}\subseteq B$. If $D\neq 0$ is locally nilpotent then $D$ has a preslice $t\in B$. For any such $t$, if we define $\alpha=D(t)$ then $B_{\alpha}=A_{\alpha}[t]=(A_{\alpha})^{[1]}$. Consequently, $\operatorname{{\rm trdeg}}_{A}(B)=1$ and $\operatorname{{\rm Frac}}(B)=(\operatorname{{\rm Frac}}(A))^{(1)}$. 3. (c)

If $D\neq 0$ is locally nilpotent and $B$ satisfies the ascending chain condition for principal ideals, then there exists an irreducible locally nilpotent derivation $\delta:B\to B$ such that $D=a\delta$ for some $a\in A$. 4. (d)

Let $S\subseteq B\setminus\{0\}$ be a multiplicative subset of $B$ containing $1$. Then $S^{-1}D:S^{-1}B\to S^{-1}B$ defined by $(S^{-1}D)(\frac{b}{s})=\frac{sD(b)-bD(s)}{s^{2}}$ is a derivation and the following hold:

1. (i)

$S^{-1}D$* is locally nilpotent if and only if $D$ is locally nilpotent and $S\subseteq A$;* 2. (ii)

if $S\subseteq A$ then $\ker S^{-1}D=S^{-1}A$ and $S^{-1}A\cap B=A$.

2.3 Definition**.**

Let $B$ be a ring of characteristic zero. If $D\in\operatorname{{\rm LND}}(B)$ then define $\deg_{D}:B\to\mathbb{N}\cup\{-\infty\}$ by declaring that $\deg_{D}(0)=-\infty$ and that \deg_{D}(x)=\max\big{\{}\,n\in\mathbb{N}\,\mid\,D^{n}(x)\neq 0\,\big{\}} for each $x\in B\setminus\{0\}$. Also define |x|_{B}=\min\big{\{}\,\deg_{D}(x)\,\mid\,D\in\operatorname{{\rm LND}}(B)\setminus\{0\}\,\big{\}} for each $x\in B\setminus\{0\}$, where we adopt the convention that $\min\varnothing=\infty$, so $|x|_{B}=\infty$ when $B$ is rigid.

3. Derivations of G-graded rings

3.1 Definition**.**

Let $(G,+)$ be an abelian group. A $G$-grading of a ring $B$ is a family $\{B_{g}\}_{g\in G}$ of subgroups of $(B,+)$ such that $B=\bigoplus_{g\in G}B_{g}$ and $B_{g}B_{h}\subseteq B_{g+h}$ for all $g,h\in G$. A $G$-graded ring is a ring $B$ together with a $G$-grading (of $B$). In the special case where $G=\mathbb{Z}$ and $B_{i}=0$ for all $i<0$, we say that $B$ is $\mathbb{N}$-graded and write $B=\bigoplus_{i\in\mathbb{N}}B_{i}$.

3.2 Definitions**.**

Suppose that $G$ is an abelian group and that $B=\bigoplus_{i\in G}B_{i}$ is a $G$-graded ring.

• •

A derivation $D:B\to B$ is homogeneous if there exists an $h\in G$ such that $D(B_{g})\subseteq B_{g+h}$ for all $g\in G$. If $D$ is homogeneous and $D\neq 0$ then $h$ is unique and we say that $D$ is homogeneous of degree $h$. The zero derivation is homogeneous of degree $-\infty$. The set of homogeneous locally nilpotent derivations of $B$ is denoted $\operatorname{{\rm HLND}}(B)$.

• •

A graded subring of $B$ is a subring $A$ of $B$ satisfying $A=\bigoplus_{g\in G}(A\cap B_{g})$. If $A$ is a graded subring of $B$ then $A$ is a $G$-graded ring ($A=\bigoplus_{g\in G}A_{g}$ is a $G$-grading of $A$, where we define $A_{g}=A\cap B_{g}$ for each $g\in G$). Note that if $D\in\operatorname{{\rm HLND}}(B)$ then $\ker D$ is a graded subring of $B$.

• •

If $A$ is a graded subring of $B$, define $G(A)$ to be the subgroup of $G$ generated by the set \big{\{}\,g\in G\,\mid\,A_{g}\neq 0\,\big{\}}.

3.3 Definition**.**

Suppose that $G$ is an abelian group and that $B=\bigoplus_{i\in G}B_{i}$ is a $G$-graded integral domain. If $S$ is a multiplicatively closed subset of $\bigcup_{i\in G}(B_{i}\setminus\{0\})$ such that $1\in S$ then the localized ring $S^{-1}B$ is a $G$-graded ring in a natural way, and if we write $S^{-1}B=R=\bigoplus_{i\in G}R_{i}$ then the subring $R_{0}$ of $S^{-1}B$ called the homogeneous localization of $B$ at $S$. Explicitly,

[TABLE]

We define $\operatorname{{\rm HFrac}}(B)$ to be the homogeneous localization of $B$ at $S=\bigcup_{i\in G}(B_{i}\setminus\{0\})$. It is clear that $\operatorname{{\rm HFrac}}(B)$ is a subfield of $\operatorname{{\rm Frac}}(B)$; we call $\operatorname{{\rm HFrac}}(B)$ the homogeneous field of fractions of $B$.

3.4 Example**.**

If $B$ is an $\mathbb{N}$-graded integral domain such that $B\neq B_{0}$ then $\operatorname{{\rm Proj}}B$ and $\operatorname{{\rm Spec}}B$ are integral schemes, $\operatorname{{\rm Frac}}(B)$ is the function field of $\operatorname{{\rm Spec}}B$, $\operatorname{{\rm HFrac}}(B)$ is the function field of $\operatorname{{\rm Proj}}B$, and $\operatorname{{\rm Frac}}(B)=(\operatorname{{\rm HFrac}}(B))^{(1)}$. (If $B=B_{0}$ then $\operatorname{{\rm Proj}}B=\varnothing$ is not an integral scheme and hence does not have a function field. Note that $\operatorname{{\rm Frac}}(B)=\operatorname{{\rm HFrac}}(B)$ whenever $B=B_{0}$.)

3.5 Definition**.**

A field extension $L/{\rm\bf k}$ is ruled if there exists a field $K$ such that ${\rm\bf k}\leq K\leq L$ and $L=K^{(1)}$.

3.6 Theorem**.**

Let $G$ be an abelian group and $B$ a $G$-graded integral domain of characteristic zero. Let $D\in\operatorname{{\rm HLND}}(B)\setminus\{0\}$, let $A=\ker D$ and suppose that $G(A)=G(B)$.

1. (a)

$\operatorname{{\rm HFrac}}(B)=(\operatorname{{\rm HFrac}}(A))^{(1)}$** 2. (b)

If ${\rm\bf k}$ is a field included in $B$ then ${\rm\bf k}\cap B_{0}$ is a field included in $\operatorname{{\rm HFrac}}(A)$. 3. (c)

If $G$ is torsion-free and ${\rm\bf k}$ is a field included in $B$ then ${\rm\bf k}\subseteq\operatorname{{\rm HFrac}}(A)$ and consequently $\operatorname{{\rm HFrac}}(B)$ is ruled over ${\rm\bf k}$.

Proof.

If $B=B_{0}$ then $\operatorname{{\rm Frac}}(B)=\operatorname{{\rm HFrac}}(B)$ and $\operatorname{{\rm Frac}}(A)=\operatorname{{\rm HFrac}}(A)$, so the Theorem follows from parts (a) and (b) of 2.2. So we may assume that $B\neq B_{0}$.

(a) Let $d=\deg D$ and $S=\bigcup_{g\in G}(A_{g}\setminus\{0\})$. Then $S^{-1}D:S^{-1}B\to S^{-1}B$ is nonzero and locally nilpotent, and also homogeneous of degree $d$. Since $-d\in G(B)$ and $G(A)=G(B)$, there exist $a,s\in S$ such that the homogeneous element $\frac{a}{s}\in S^{-1}A$ has degree $-d$. Then $\frac{a}{s}S^{-1}D:S^{-1}B\to S^{-1}B$ is a homogeneous derivation of degree [math]; since $\frac{a}{s}\in\ker(S^{-1}D)$ and $S^{-1}D$ is locally nilpotent, $\frac{a}{s}S^{-1}D$ is locally nilpotent. Let $B_{(S)}$ denote the degree 0 subring of $S^{-1}B$ and let $D_{(S)}:B_{(S)}\to B_{(S)}$ be the restriction of $\frac{a}{s}S^{-1}D$. Then $D_{(S)}$ is a locally nilpotent derivation whose kernel is B_{(S)}\cap\ker\big{(}\frac{a}{s}S^{-1}D\big{)}=B_{(S)}\cap S^{-1}A=A_{(S)}=\operatorname{{\rm HFrac}}(A).

We show that $D_{(S)}$ is nonzero. It is straightforward to verify that $D$ has a homogeneous preslice $t\in B$. Since $G(A)=G(B)$, there exist $a^{\prime},s^{\prime}\in S$ such that $\deg(\frac{a^{\prime}}{s^{\prime}})=-\deg(t)$. Then $\frac{a^{\prime}t}{s^{\prime}}\in B_{(S)}$ and, since $\frac{a^{\prime}}{s^{\prime}}\in\ker S^{-1}D$, D_{(S)}(\frac{a^{\prime}t}{s^{\prime}})=\big{(}\frac{a}{s}S^{-1}D\big{)}(\frac{a^{\prime}t}{s^{\prime}})=\frac{a^{\prime}}{s^{\prime}}\frac{a}{s}S^{-1}D(t)\neq 0.

Since $D_{(S)}\neq 0$ and $\ker D_{(S)}=\operatorname{{\rm HFrac}}(A)$ is a field, 2.2(b) implies that $B_{(S)}=(\operatorname{{\rm HFrac}}(A))^{[1]}$. Using $G(A)=G(B)$ once again, we see that $\operatorname{{\rm HFrac}}(B)$ is the field of fractions of $B_{(S)}$; so $\operatorname{{\rm HFrac}}(B)=(\operatorname{{\rm HFrac}}(A))^{(1)}$. This proves (a).

(b) If ${\rm\bf k}$ is a field included in $B$ then it is clear that ${\rm\bf k}\cap B_{0}$ is a field and 2.2(a) implies that ${\rm\bf k}\subseteq A$, so ${\rm\bf k}\cap B_{0}\subseteq A\cap B_{0}=A_{0}\subseteq\operatorname{{\rm HFrac}}(A)$.

(c) It is well known that if $G$ is a torsion-free abelian group and $B$ is a $G$-graded domain then any field included in $B$ is in fact included in $B_{0}$ (see [2, Lemma 2.4.7], for instance). So in the present situation we have ${\rm\bf k}\subseteq B_{0}$; then (b) implies that ${\rm\bf k}\subseteq\operatorname{{\rm HFrac}}(A)$ and (a) implies that $\operatorname{{\rm HFrac}}(B)$ is ruled over ${\rm\bf k}$. ∎

3.7 Remark**.**

In the special case where the grading is an $\mathbb{N}$-grading and is nontrivial ($B\neq B_{0}$), Theorem 3.6(c) asserts that the function field of $\operatorname{{\rm Proj}}B$ is ruled over ${\rm\bf k}$. (See Example 3.4.) The same remark applies to Proposition 3.10, below.

3.8****Homogenization.

Let ${\rm\bf k}$ be a field of characteristic zero and $B=\bigoplus_{i\in\mathbb{Z}}B_{i}$ a $\mathbb{Z}$-graded affine ${\rm\bf k}$-domain. One can show that if $D\in\operatorname{{\rm LND}}(B)$ then the subset S_{D}\overset{\text{\tiny\rm def}}{=}\big{\{}\,\deg(Dx)-\deg(x)\,\mid\,x\in B\setminus\{0\}\,\big{\}} of $\mathbb{Z}\cup\{-\infty\}$ has a greatest element (where $\deg:B\to\mathbb{Z}\cup\{-\infty\}$ is the degree function determined by the grading of $B$); one defines $\deg(D)=\max(S_{D})\in\mathbb{Z}\cup\{-\infty\}$. Clearly, $\deg(D)=-\infty$ $\Leftrightarrow$ $D=0$. Also, if $D$ happens to be homogeneous then $\deg(D)$ coincides with the usual degree of a homogeneous derivation (Definition 3.2).

If $D\in\operatorname{{\rm LND}}(B)$ then there is a natural way to define an element $\tilde{D}$ of $\operatorname{{\rm HLND}}(B)$ satisfying (in particular) $\deg(\tilde{D})=\deg(D)$. Indeed, if $D=0$ then set $\tilde{D}=0$. If $D\in\operatorname{{\rm LND}}(B)\setminus\{0\}$ then let $d=\deg(D)\in\mathbb{Z}$ and define (for each $i\in\mathbb{Z}$) a map $\tilde{D}_{i}:B_{i}\to B_{i+d}$ by $\tilde{D}_{i}(x)=p_{i+d}(Dx)$, where $p_{j}:B\to B_{j}$ is the canonical projection; then if $b=\sum_{i\in\mathbb{Z}}b_{i}\in B$ ($b_{i}\in B_{i}$ for all $i$, $b_{i}=0$ for almost all $i$), define $\tilde{D}(b)=\sum_{i}\tilde{D}_{i}(b_{i})$; one can check that $\tilde{D}\in\operatorname{{\rm HLND}}(B)\setminus\{0\}$ and $\deg(\tilde{D})=\deg(D)$. Note in particular that $D\neq 0\implies\tilde{D}\neq 0$. The derivation $\tilde{D}$ is sometimes called the homogenization of $D$. This is in fact a special case of the process of replacing $D:B\to B$ by $\operatorname{{\rm\bf Gr}}(D):\operatorname{{\rm\bf Gr}}(B)\to\operatorname{{\rm\bf Gr}}(B)$, so one can also call $\tilde{D}$ the associated homogeneous derivation of $D$. Note the following consequence of the above discussion:

Let ${\rm\bf k}$ be a field of characteristic zero and $B$ a $\mathbb{Z}$-graded affine ${\rm\bf k}$-domain. If $B$ is not rigid then $\operatorname{{\rm HLND}}(B)\neq\{0\}$.

Refer to [4] for proofs of the claims made in 3.8, and for an in-depth treatment of this topic.

3.9 Definitions**.**

Let $n\geq 2$ and $S=(a_{1},a_{2},\dots,a_{n})\in\mathbb{Z}^{n}$.

• •

Define111By convention, $\gcd(S)\geq 0$ and $\operatorname{{\rm lcm}}(S)\geq 0$. $\gcd(S)=\gcd(a_{1},\dots,a_{n})$ and $\operatorname{{\rm lcm}}(S)=\operatorname{{\rm lcm}}(a_{1},\dots,a_{n})$.

• •

If $\gcd(S)=1$, we say that $S$ is normal. If $S\neq(0,\dots,0)$ then the tuple $S^{\prime}=(\frac{a_{1}}{d},\dots,\frac{a_{n}}{d})$ (where $d=\gcd(S)$) is normal, and is called the normalization of $S$.

• •

For each $j\in\{1,\dots,n\}$, define $S_{j}=(a_{1},\dots\,\widehat{a_{j}}\,\dots,a_{n})$ ($j$-th component omitted). More generally, given a proper subset $J$ of $\{1,\dots,n\}$ we define $S_{J}=(a_{i_{1}},\dots,a_{i_{s}})$, where

$i_{1}<\cdots<i_{s}$ are the elements of $\{1,\dots,n\}\setminus J$.

• •

We define the sets

[TABLE]

Then we define the type and the cotype of $S$ by:

[TABLE]

Note that $\operatorname{{\rm type}}(S),\operatorname{{\rm cotype}}(S)\in\{0,1,\dots,n\}$ and that, if $S^{\prime}$ is the normalization of $S$, then $\operatorname{{\rm type}}(S)=\operatorname{{\rm type}}(S^{\prime})$ and $\operatorname{{\rm cotype}}(S)=\operatorname{{\rm cotype}}(S^{\prime})$.

3.10 Proposition**.**

Let ${\rm\bf k}$ be a field of characteristic zero and $B$ a $\mathbb{Z}$-graded ${\rm\bf k}$-domain. Suppose that there exist homogeneous prime elements $x_{1},\dots,x_{n}$ $(n\geq 2)$ of $B$ satisfying the following conditions, where $d_{i}=\deg x_{i}$ for all $i$:

• •

$\langle d_{1},\dots,d_{n}\rangle=\mathbb{Z}(B)$* and $\operatorname{{\rm type}}(d_{1},\dots,d_{n})=0$*

• •

for any choice of distinct $i,j$, the elements $x_{i},x_{j}$ are not associates in $B$.

Then the following hold.

1. (a)

$\mathbb{Z}(\ker D)=\mathbb{Z}(B)$* for all $D\in\operatorname{{\rm HLND}}(B)$.* 2. (b)

If $B$ is ${\rm\bf k}$-affine222We mean that $B$ is finitely generated as a ${\rm\bf k}$-algebra.* and not rigid then $\operatorname{{\rm HFrac}}(B)$ is ruled over ${\rm\bf k}$.*

Proof.

The assumption $\langle d_{1},\dots,d_{n}\rangle=\mathbb{Z}(B)$ and $\operatorname{{\rm type}}(d_{1},\dots,d_{n})=0$ is equivalent to:

for each subset $I\subset\{1,\dots,n\}$ of cardinality $n-1$, \big{\{}\,\deg(x_{i})\,\mid\,i\in I\,\big{\}} generates $\mathbb{Z}(B)$.

So assertion (a) is the special case $G=\mathbb{Z}$ of [5, Corollary 4.2]. To prove (b), suppose that $B$ is ${\rm\bf k}$-affine and not rigid; then 3.8 implies that $\operatorname{{\rm HLND}}(B)\setminus\{0\}\neq\varnothing$. Pick $D\in\operatorname{{\rm HLND}}(B)\setminus\{0\}$, then $\mathbb{Z}(\ker D)=\mathbb{Z}(B)$ by part (a), so Theorem 3.6(c) implies that $\operatorname{{\rm HFrac}}(B)$ is ruled over ${\rm\bf k}$. ∎

4. Rigidity of Pham-Brieskorn Rings

Let ${\rm\bf k}$ be a field of characteristic zero.

4.1 Definition**.**

Given $n\geq 3$ and $S=(a_{1},\dots,a_{n})\in(\mathbb{N}\setminus\{0\})^{n}$, we use the notation

[TABLE]

The ${\rm\bf k}$-domain $B_{a_{1},\dots,a_{n}}$ is called a Pham-Brieskorn ring. We use the capital letters $X_{i}$ to represent the variables in the polynomial ring, and define $x_{i}=\pi(X_{i})\in B_{S}$ where $\pi:{\rm\bf k}[X_{1},\dots,X_{n}]\to B_{S}$ is the canonical quotient map. Thus $B_{S}={\rm\bf k}[x_{1},\dots,x_{n}]$. Let (d_{1},\dots,d_{n})=\big{(}\frac{L}{a_{1}},\dots,\frac{L}{a_{n}}\big{)}, where $L=\operatorname{{\rm lcm}}(S)$; then there is a unique $\mathbb{N}$-grading of $B_{S}$ with the property that $x_{i}$ is homogeneous of degree $d_{i}$ for all $i=1,\dots,n$. We call it the standard $\mathbb{N}$-grading of $B_{S}$.

4.2 Lemma**.**

Given $n\geq 2$ and $S=(a_{1},\dots,a_{n})\in(\mathbb{N}\setminus\{0\})^{n}$, define \bar{S}=\big{(}\frac{L}{a_{1}},\dots,\frac{L}{a_{n}}\big{)} where $L=\operatorname{{\rm lcm}}(S)$. Then the following hold.

1. (a)

$\bar{S}$* is normal and $\overline{(\bar{S})}$ is the normalization of $S$.* 2. (b)

$J^{*}(\bar{S})=J(S)$*   and  $J^{*}(S)=J(\bar{S})$.* 3. (c)

$\operatorname{{\rm type}}(\bar{S})=\operatorname{{\rm cotype}}(S)$*  and  $\operatorname{{\rm type}}(S)=\operatorname{{\rm cotype}}(\bar{S})$.*

The proof of Lemma 4.2 is left to the reader. We deduce the following useful triviality:

4.3 Lemma**.**

Let $n\geq 4$ and $S=(a_{1},\dots,a_{n})\in(\mathbb{N}\setminus\{0\})^{n}$, and consider $B_{S}$, $x_{1},\dots,x_{n}$ and $(d_{1},\dots,d_{n})$ as in Definition 4.1.

1. (a)

$\gcd(d_{1},\dots,d_{n})=1$*   and  $\operatorname{{\rm type}}(d_{1},\dots,d_{n})=\operatorname{{\rm cotype}}(a_{1},\dots,a_{n})$* 2. (b)

For each $i\in\{1,\dots,n\}$, we have

[TABLE] 3. (c)

$x_{1},\dots,x_{n}$* are homogeneous prime elements of $B_{S}$ and are pairwise non-associates.*

Proof.

Since $(d_{1},\dots,d_{n})=\bar{S}$, assertions (a) and (b) follow from Lemma 4.2. For each $i$, define $S_{i}$ as in Definition 3.9; then $B_{S}/\langle x_{i}\rangle\cong B_{S_{i}}$, which is a domain because $n\geq 4$. So $x_{1},\dots,x_{n}$ are homogeneous prime elements of $B_{S}$, and it is clear that they are pairwise non-associates. ∎

Recall that a ${\rm\bf k}$-variety $X$ is said to be ruled if the function field of $X$ is ruled over ${\rm\bf k}$, in the sense of Definition 3.5.

4.4 Theorem**.**

Let $n\geq 4$, and let $S\in(\mathbb{N}\setminus\{0\})^{n}$ be such that $\operatorname{{\rm cotype}}(S)=0$.

1. (a)

We have $\mathbb{Z}(\ker D)=\mathbb{Z}$ for all $D\in\operatorname{{\rm HLND}}(B_{S})$. 2. (b)

If $B_{S}$ is not rigid, then the projective ${\rm\bf k}$-variety $\operatorname{{\rm Proj}}B_{S}$ is ruled over ${\rm\bf k}$.

Proof.

By Example 3.4, the function field of $\operatorname{{\rm Proj}}B_{S}$ is $\operatorname{{\rm HFrac}}(B_{S})$. Consider $x_{1},\dots,x_{n}$ and $(d_{1},\dots,d_{n})$ as in Definition 4.1. By Lemma 4.3, $x_{1},\dots,x_{n}$ satisfy the hypothesis of Proposition 3.10 (in particular $\operatorname{{\rm type}}(d_{1},\dots,d_{n})=\operatorname{{\rm cotype}}(S)=0$). So Proposition 3.10 implies that (a) and (b) hold. ∎

4.5 Remark**.**

If $a\geq n\geq 4$ then the Fermat variety F_{a,n}=\operatorname{{\rm Proj}}\big{(}{\rm\bf k}[X_{1},\dots,X_{n}]/\langle X_{1}^{a}+\cdots+X_{n}^{a}\rangle\big{)} is not uniruled, hence not ruled. (Over a field of characteristic zero, a smooth hypersurface of degree $d$ in $\mathbb{P}^{N}$ is uniruled if and only if $d\leq N$.)

4.6 Corollary**.**

If $S=(\underbrace{a,\dots,a}_{n})$ satisfies $a\geq n\geq 4$, then $B_{S}$ is rigid.

Proof.

Note that $\operatorname{{\rm cotype}}(S)=0$. If $B_{S}$ is not rigid then (by Theorem 4.4) $\operatorname{{\rm Proj}}B_{S}=F_{a,n}$ is ruled, which is not the case by Remark 4.5. ∎

We shall now develop a different approach for proving that $B_{S}$ is rigid in certain cases. We need the following result, which is part (b) of Corollary 3.3 of [10]. See 2.3 for the definition of $|u|_{A}$.

4.7 Corollary**.**

Suppose $R$ is a $\mathbb{Z}$-graded affine ${\rm\bf k}$-domain, $f\in R$ is homogeneous, and $n\geq 2$ is an integer not dividing $\deg f$. Set $g=\gcd(n,\deg f)$, define the rings

[TABLE]

and assume that $B$ is a domain. Then $A$ is a domain and

[TABLE]

4.8 Definition**.**

Let $n\geq 3$.

1. (a)

Given $S=(a_{1},\dots,a_{n})\in(\mathbb{N}\setminus\{0\})^{n}$ and $i\in\{1,\dots,n\}$, define $g_{i}(S)=\gcd(a_{i},\operatorname{{\rm lcm}}(S_{i}))$ (recalling the definition of $S_{i}$ from Definition 3.9). 2. (b)

Let $S=(a_{1},\dots,a_{n})$ and $S^{\prime}=(a_{1}^{\prime},\dots,a_{n}^{\prime})$ be elements of $(\mathbb{N}\setminus\{0\})^{n}$ and let $i\in\{1,\dots,n\}$. We write $S\leq^{i}S^{\prime}$ if and only if

[TABLE]

We write $S<^{i}S^{\prime}$ if and only if $S\leq^{i}S^{\prime}$ and $S\neq S^{\prime}$.

Observe that $\leq^{i}$ is a partial order on $(\mathbb{N}\setminus\{0\})^{n}$ (transitivity follows from the fact that if $S\leq^{i}S^{\prime}$ then $g_{i}(S)=g_{i}(S^{\prime})$).

4.9 Proposition**.**

Let $n\geq 3$.

1. (a)

Suppose that $S^{*},S\in(\mathbb{N}\setminus\{0\})^{n}$ and $i\in\{1,\dots,n\}$ satisfy $S^{*}\leq^{i}S$. If $B_{S^{*}}$ is rigid then so is $B_{S}$. 2. (b)

Let $S,S^{\prime}\in(\mathbb{N}\setminus\{0\})^{n}$, and suppose that there exist $i\in\{1,\dots,n\}$ and $S^{*}\in(\mathbb{N}\setminus\{0\})^{n}$ such that $S^{*}<^{i}S$ and $S^{*}<^{i}S^{\prime}$. Then $B_{S}$ is rigid if and only if $B_{S^{\prime}}$ is rigid.

Proof.

(b) We may assume that $i=n$. Consider $S=(a_{1},\dots,a_{n})$, $S^{\prime}=(a_{1}^{\prime},\dots,a_{n}^{\prime})$ and $S^{*}=(a_{1}^{*},\dots,a_{n}^{*})$ satisfying $S^{*}<^{n}S$ and $S^{*}<^{n}S^{\prime}$. The hypothesis implies that

[TABLE]

where $g=g_{n}(S)=g_{n}(S^{*})=g_{n}(S^{\prime})$. Let $L=\operatorname{{\rm lcm}}(a_{1},\dots,a_{n-1})$ and define an $\mathbb{N}$-grading on $R={\rm\bf k}[X_{1},\dots,X_{n-1}]$ by declaring that (for each $i=1,\dots,n-1$) $X_{i}$ is homogeneous of degree $\frac{a_{n}^{*}}{g}\frac{L}{a_{i}}$. Then $f=X_{1}^{a_{1}}+\cdots+X_{n-1}^{a_{n-1}}$ is homogeneous of degree $\frac{a_{n}^{*}L}{g}$. Consider the rings

[TABLE]

Observe that $A\cong B_{S^{*}}$, $B\cong B_{S}$ and $B^{\prime}\cong B_{S^{\prime}}$. The reader may verify that $a_{n}\nmid\deg f$, $a_{n}^{\prime}\nmid\deg f$ and $\gcd(a_{n},\deg f)=a_{n}^{*}=\gcd(a_{n}^{\prime},\deg f)$; thus each one of the pairs $(A,B)$, $(A,B^{\prime})$ satisfies the hypothesis of Corollary 4.7. By that result, $B$ is rigid $\Leftrightarrow$ $|u|_{A}\geq 2$ $\Leftrightarrow$ $B^{\prime}$ is rigid, so (b) is proved. It also follows that (a) is true, because if $A$ is rigid then $|u|_{A}=\infty\geq 2$, so $B$ is rigid. ∎

4.10 Corollary**.**

* ***

1. (a)

If $k_{1},k_{2},k_{3},k_{4}$ are pairwise relatively prime positive integers and $a\geq 3$ then $B_{ak_{1},ak_{2},ak_{3},ak_{4}}$ is rigid. 2. (b)

If $k_{1},\dots,k_{n}$ are pairwise relatively prime positive integers and $a\geq n\geq 4$ then $B_{ak_{1},\dots,ak_{n}}$ is rigid.

Proof.

Define $S^{0}=(a,a,a,a)$ to prove (a), and $S^{0}=(\overbrace{a,\dots,a}^{n})$ to prove (b). Note that $B_{S^{0}}$ is rigid (by Theorem 1.3(c) if $S^{0}=(3,3,3,3)$, by Corollary 4.6 in the other cases). Define $S^{1}=(ak_{1},a,a,\dots)$, $S^{2}=(ak_{1},ak_{2},a,\dots)$, …, $S^{n-1}=(ak_{1},\dots,ak_{n-1},a)$, $S^{n}=(ak_{1},\dots,ak_{n})$; then $S^{0}\leq^{1}S^{1}\leq^{2}S^{2}\leq^{3}\cdots\leq^{n-1}S^{n-1}\leq^{n}S^{n}$. Since $B_{S^{0}}$ is rigid, so is $B_{S^{n}}$ by Proposition 4.9. ∎

4.11 Lemma**.**

Given $n\geq 3$ and $S=(a_{1},\dots,a_{n})\in(\mathbb{N}\setminus\{0\})^{n}$,

[TABLE]

Proof.

If $j\in J(S)$ then $a_{j}\nmid\operatorname{{\rm lcm}}(S_{j})$, so $g_{j}(S)\neq a_{j}$ and hence

[TABLE]

Conversely, if $(a_{1}^{\prime},\dots,a_{n}^{\prime})<^{j}S$ then $g_{j}(S)\mid a_{j}^{\prime}\mid a_{j}$ and $a_{j}^{\prime}\neq a_{j}$, so $j\in J(S)$. ∎

4.12 Proposition**.**

Let $n\geq 3$, $S=(a_{1},\dots,a_{n})\in(\mathbb{N}\setminus\{0\})^{n}$ and

[TABLE]

If $\sum\limits_{i\in I(S)}\frac{1}{a_{i}}<\frac{1}{n-2}$ then $B_{S}$ is rigid.

Proof.

Let us write $I=I(S)$ and $J=J(S)$. If $J=\varnothing$ then $\sum\limits_{i=1}^{n}\frac{1}{a_{i}}<\frac{1}{n-2}$ so we are done by Theorem 1.2. Assume that $J\neq\varnothing$ and let $j_{1}<\cdots<j_{k}$ be the elements of $J$. Choose distinct primes $p_{j_{1}},\dots,p_{j_{k}}$ such that $\gcd(\prod_{j\in J}p_{j},\,\prod_{i=1}^{n}a_{i})=1$. Choose positive integers $e_{j_{1}},\dots,e_{j_{k}}$ such that $\sum\limits_{i\in I}\frac{1}{a_{i}}+\sum\limits_{j\in J}\frac{1}{a_{j}p_{j}^{e_{j}}}<\frac{1}{n-2}$. We inductively define a sequence $S^{0},S^{1},\dots,S^{k}$ of elements of $(\mathbb{N}\setminus\{0\})^{n}$ by setting $S^{0}=S$ and, for each $\nu=1,\dots,k$,

[TABLE]

where the $p_{j_{\nu}}^{e_{j_{\nu}}}$ is in the $j_{\nu}$-th position. The reader can check that $J(S^{\nu})=J(S)=J$ for all $\nu$ and that $S^{0}\leq^{j_{1}}S^{1}\leq^{j_{2}}S^{2}\leq^{j_{3}}\cdots\leq^{j_{k}}S^{k}$.

Let $\nu\in\{1,\dots,k\}$; since $j_{\nu}\in J=J(S^{\nu-1})$, Lemma 4.11 implies that $S^{\nu-1}$ is not minimal with respect to $\leq^{j_{\nu}}$, i.e., there exists $S^{\nu-1}_{*}\in(\mathbb{N}\setminus\{0\})^{n}$ satisfying $S^{\nu-1}_{*}<^{j_{\nu}}S^{\nu-1}<^{j_{\nu}}S^{\nu}$; then Proposition 4.9(b) implies that $B_{S^{\nu-1}}$ is rigid if and only if $B_{S^{\nu}}$ is rigid. As this holds for all $\nu\in\{1,\dots,k\}$, $B_{S}=B_{S^{0}}$ is rigid if and only if $B_{S^{k}}$ is rigid. Now $B_{S^{k}}$ is indeed rigid by Theorem 1.2, because if we write $S^{k}=(a_{1}^{\prime},\dots,a_{n}^{\prime})$ then $\sum\limits_{i=1}^{n}\frac{1}{a_{i}^{\prime}}=\sum\limits_{i\in I}\frac{1}{a_{i}}+\sum\limits_{j\in J}\frac{1}{a_{j}p_{j}^{e_{j}}}<\frac{1}{n-2}$. So $B_{S}$ is rigid. ∎

Proposition 4.12 shows that many Pham-Brieskorn varieties that are not shown to be rigid by Theorems 1.1–1.3 are indeed rigid. In the case $n=4$, the following is an immediate consequence of Proposition 4.12:

4.13 Corollary**.**

If $a,b,c,d\in\mathbb{N}\setminus\{0\}$ satisfy $a\nmid\operatorname{{\rm lcm}}(b,c,d)$ and $\frac{1}{b}+\frac{1}{c}+\frac{1}{d}<\frac{1}{2}$ then $B_{a,b,c,d}$ is rigid.

We need the following:

4.14 Corollary**.**

Let $n\geq 4$ and $S\in(\mathbb{N}\setminus\{0\})^{n}$. Consider $B_{S}={\rm\bf k}[x_{1},\dots,x_{n}]$ with $x_{1},\dots,x_{n}$ as in Definition 4.1. Then for every $D\in\operatorname{{\rm HLND}}(B_{S})$ the following conditions hold.

1. (a)

If $j\in J(S)$ then $D^{2}(x_{j})=0$. 2. (b)

If $j_{1},j_{2}$ are distinct elements of $J(S)$ then $D(x_{j_{1}})=0$ or $D(x_{j_{2}})=0$.

Proof.

In view of Lemma 4.3(b), this is a special case of Corollary 6.3 of [5]. ∎

4.15 Proposition**.**

Let $n\geq 4$ and $S\in(\mathbb{N}\setminus\{0\})^{n}$. Assume that $J(S)\neq\varnothing$ and that

[TABLE]

Then $B_{S}$ is rigid.

Proof.

If $|J(S)|=1$ then $\min(|J(S)|-1,n-3)=0$, so $(*)$ reads “$B_{S}$ is rigid,” so the Proposition is trivially true in this case. From now-on, we assume that $|J(S)|>1$. By contradiction, assume that $B_{S}$ is not rigid. Then $\operatorname{{\rm HLND}}(B_{S})\neq\{0\}$ by 3.8; choose an irreducible $D\in\operatorname{{\rm HLND}}(B_{S})\setminus\{0\}$ (this is possible by 2.2(c)). By Corollary 4.14, we have $D(x_{j_{1}})=0$ or $D(x_{j_{2}})=0$ for every choice of distinct $j_{1},j_{2}\in J(S)$. So there exists a subset $J^{\prime}$ of $J(S)$ such that $|J^{\prime}|=|J(S)|-1$ and $D(x_{j})=0$ for all $j\in J^{\prime}$; hence there exists a $J\subseteq J^{\prime}\subset J(S)$ such that $|J|=\min(|J(S)|-1,n-3)$ and $D(x_{j})=0$ for all $j\in J$. We might as well assume that $J=\{m+1,m+2,\dots,n\}$ for some $m\geq 3$. For each $i$ such that $m\leq i\leq n$ we define $P(i)$ to be the statement

there exists an irreducible $D_{i}\in\operatorname{{\rm HLND}}(B_{a_{1},\dots,a_{i}})\setminus\{0\}$

satisfying $D_{i}(x_{j})=0$ for all $j\in\{1,\dots,i\}\cap J$

where $x_{j}$ denotes the image of $X_{j}$ in $B_{a_{1},\dots,a_{i}}$. Then $P(n)$ is true, with $D_{n}=D$. Assume that $i$ is such that $m<i\leq n$ and such that $P(i)$ is true. Note that $i\in J$, so $D_{i}(x_{i})=0$, so $D_{i}$ induces a homogeneous locally nilpotent derivation $\bar{D}_{i}$ of the ring $B_{a_{1},\dots,a_{i}}/\langle x_{i}\rangle\cong B_{a_{1},\dots,a_{i-1}}$, and $\bar{D}_{i}\neq 0$ because $D_{i}$ is irreducible. Moreover, $\bar{D}_{i}(\bar{x}_{j})=0$ for all $j\in\{1,\dots,i-1\}\cap J$. By 2.2(c), there exists an irreducible $D_{i-1}\in\operatorname{{\rm HLND}}(B_{a_{1},\dots,a_{i-1}})\setminus\{0\}$ such that $\ker D_{i-1}=\ker\bar{D}_{i}$, so $P(i-1)$ is true. By descending induction, $P(n),P(n-1),\dots,P(m)$ are all true. Since $P(m)$ is true and $S_{J}=(a_{1},\dots,a_{m})$, we have $D_{m}\in\operatorname{{\rm HLND}}(B_{S_{J}})\setminus\{0\}$, so $B_{S_{J}}$ is not rigid, contradicting $(*)$. ∎

We generalize Theorem 1.3(e):

4.16 Corollary**.**

Let $n\geq 4$ and $S\in T_{n}$. If $\operatorname{{\rm cotype}}(S)\geq n-2$, then $B_{S}$ is rigid.

Proof.

We have $|J(S)|=\operatorname{{\rm cotype}}(S)\geq n-2$, so $\min(|J(S)|-1,n-3)=n-3$. Let $J$ be any subset of $J(S)$ satisfying $|J|=\min(|J(S)|-1,n-3)=n-3$; since $S\in T_{n}$, we have $S_{J}\in T_{3}$, so $B_{S_{J}}$ is rigid by Theorem 1.1. Then Proposition 4.15 implies that $B_{S}$ is rigid. ∎

4.17 Example**.**

Proposition 4.15 can settle cases not covered by Corollary 4.16 or by any of the previous results. For instance:

If $a_{1},\dots,a_{m}\in(\mathbb{N}\setminus\{0\})^{m}$ ($m\geq 1$) satisfy $a_{i}\nmid\operatorname{{\rm lcm}}(3,a_{1},\dots\widehat{a_{i}}\dots,a_{m})$ for all $i\in\{1,\dots,m\}$, then $B_{a_{1},\dots,a_{m},3,3,3}$ is rigid.

Indeed, let $S=(a_{1},\dots,a_{m},3,3,3)$, then $J(S)=\{1,\dots,m\}$ and (by Theorem 1.3(c)) $B_{a_{i},3,3,3}$ is rigid for each $i\in J(S)$, so $B_{S}$ is rigid by Proposition 4.15. For instance, $B_{2,5,7,3,3,3}$ is rigid.

4.18 Notation**.**

Given $S=(a_{1},\dots,a_{n})\in(\mathbb{N}\setminus\{0\})^{n}$ ($n\geq 2$) and a subset $M$ of $\{1,\dots,n\}$, define a positive integer $\Delta_{M}(S)$ by $\Delta_{\varnothing}(S)=1$ and

[TABLE]

Observe that

[TABLE]

because $\operatorname{{\rm lcm}}(S_{j})=\operatorname{{\rm lcm}}(S)$ for each $j\in\{1,\dots,n\}\setminus J(S)$, and $\operatorname{{\rm lcm}}(S_{j})$ is a proper divisor of $\operatorname{{\rm lcm}}(S)$ for each $j\in J(S)$.

4.19 Proposition**.**

Let $n\geq 4$, $S\in(\mathbb{N}\setminus\{0\})^{n}$ and M=\big{\{}\,j\in J(S)\,\mid\,\text{B_{S_{j}} is rigid}\,\big{\}}.

1. (a)

$\mathbb{Z}(\ker D)\subseteq\Delta_{M}(S)\cdot\mathbb{Z}$*   for all $D\in\operatorname{{\rm HLND}}(B_{S})\setminus\{0\}$.* 2. (b)

$\Delta_{M}(S)=1\iff M=\varnothing$**

Proof.

We use the following notation: $S=(a_{1},\dots,a_{n})$, $\overline{S}=(d_{1},\dots\,d_{n})$, $x_{1},\dots,x_{n}$ as in Definition 4.1, $L=\operatorname{{\rm lcm}}(S)$ and, for each $i\in\{1,\dots,n\}$, $L_{i}=\operatorname{{\rm lcm}}(S_{i})$ and $\delta_{i}=\gcd(d_{1},\dots\,\widehat{d_{i}}\,\dots,d_{n})$. Then $\delta_{i}=\gcd(\frac{L}{a_{1}},\dots\,\widehat{\textstyle\frac{L}{a_{i}}}\,\dots,\frac{L}{a_{n}})=\frac{L}{L_{i}}\gcd(\frac{L_{i}}{a_{1}},\dots\,\widehat{\frac{L_{i}}{a_{i}}}\,\dots,\frac{L_{i}}{a_{n}})=\frac{L}{L_{i}}$ for each $i\in\{1,\dots,n\}$, so $\operatorname{{\rm lcm}}(\delta_{j}\,:\,j\in M)=\operatorname{{\rm lcm}}(\frac{L}{L_{j}}\,:\,j\in M)=\frac{L}{\gcd(L_{j}\,:\,j\in M)}=\Delta_{M}(S)$ and consequently $\bigcap_{j\in M}\delta_{j}\mathbb{Z}=\Delta_{M}(S)\cdot\mathbb{Z}$. So, to prove (a), it suffices to show that

[TABLE]

We prove this by contradiction: suppose that $D\in\operatorname{{\rm HLND}}(B_{S})\setminus\{0\}$ and $j\in M$ satisfy $\mathbb{Z}(\ker D)\nsubseteq\delta_{j}\mathbb{Z}$. By 2.2(c), there exists an irreducible $D^{\prime}\in\operatorname{{\rm HLND}}(B_{S})\setminus\{0\}$ with $\ker(D^{\prime})=\ker(D)$, so we may choose $D$ irreducible. Then there exists a homogeneous element $h\in\ker(D)\setminus\{0\}$ such that $\deg(h)\notin\delta_{j}\mathbb{Z}$. We can write $h=\sum_{(i_{1},\dots,i_{n})\in E}\lambda_{i_{1},\dots,i_{n}}x_{1}^{i_{1}}\cdots x_{n}^{i_{n}}$ for some finite set $E\subset\mathbb{N}^{n}$, where for each $(i_{1},\dots,i_{n})\in E$ we have $\lambda_{i_{1},\dots,i_{n}}\in{\rm\bf k}^{*}$ and $\deg(x_{1}^{i_{1}}\cdots x_{n}^{i_{n}})=\deg(h)\notin\delta_{j}\mathbb{Z}$. So for each $(i_{1},\dots,i_{n})\in E$ we have $\deg(x_{j}^{i_{j}})\notin\delta_{j}\mathbb{Z}$ and in particular $i_{j}>0$. This shows that $x_{j}\mid h$ in $B_{S}$. Since $\ker(D)$ is factorially closed in $B_{S}$, we get $x_{j}\in\ker D$. Then $D$ induces a locally nilpotent derivation $\bar{D}$ of the ring $B_{S}/\langle x_{j}\rangle\cong B_{S_{j}}$, and $\bar{D}\neq 0$ because $D(B_{S})\nsubseteq\langle x_{j}\rangle$ (since $D$ is irreducible). So $B_{S_{j}}$ is not rigid, which contradicts $j\in M$. This proves (1), so assertion (a) is proved. Part (b) follows from the observation (made in 4.18) that, for any subset $M$ of $\{1,\dots,n\}$, we have $\Delta_{M}(S)=1$ if and only if $M\cap J(S)=\varnothing$. ∎

It immediately follows:

4.20 Corollary**.**

Let $n\geq 4$ and $S\in(\mathbb{N}\setminus\{0\})^{n}$. If some $D\in\operatorname{{\rm HLND}}(B_{S})\setminus\{0\}$ satisfies $\mathbb{Z}(\ker D)=\mathbb{Z}$ then $B_{S_{j}}$ is non-rigid for every $j\in J(S)$.

Here is another consequence of Proposition 4.19, valid for $n=4$:

4.21 Corollary**.**

Let $S\in T_{4}$ be such that $\operatorname{{\rm cotype}}(S)>0$ and define $\delta=\Delta_{J(S)}(S)\in\mathbb{Z}$. Then

[TABLE]

Proof.

We have $J(S)\neq\varnothing$ because $\operatorname{{\rm cotype}}(S)>0$. If $j\in J(S)$ then $S_{j}\in T_{3}$, so $B_{S_{j}}$ is rigid by Theorem 1.1. So the set M=\big{\{}\,j\in J(S)\,\mid\,\text{B_{S_{j}} is rigid}\,\big{\}} of Proposition 4.19 is equal to $J(S)$ (which is not empty). The desired conclusion follows from Proposition 4.19. ∎

5. A remark about $\operatorname{{\rm Proj}}(B_{S})$

As in the preceding section, we assume that ${\rm\bf k}$ is a field of characteristic zero.

5.1 Proposition** (Exercise 9.5 in [7]).**

Let $R=R_{0}\oplus R_{1}\oplus R_{2}\oplus\dots$ be an $\mathbb{N}$-graded Noetherian ring, let $d>0$ and let $R^{(d)}=R_{0}\oplus R_{d}\oplus R_{2d}\oplus\dots$. Then $\operatorname{{\rm Proj}}R\cong\operatorname{{\rm Proj}}R^{(d)}$.

5.2 Proposition**.**

Let $S=(a_{1},\dots,a_{n})$, $S^{\prime}=(a_{1}^{\prime},\dots,a_{n}^{\prime})\in(\mathbb{N}\setminus\{0\})^{n}$ ($n\geq 3$) and assume that $i\in\{1,\dots,n\}$ is such that $S\leq^{i}S^{\prime}$. Then $B_{S}\cong B_{S^{\prime}}^{(k)}$, where we define $k=a_{i}^{\prime}/a_{i}\in\mathbb{N}\setminus\{0\}$. Consequently, $\operatorname{{\rm Proj}}B_{S}\cong\operatorname{{\rm Proj}}B_{S^{\prime}}$.

Proof.

We may assume that $i=1$. Define $(d_{1},\dots,d_{n})=\overline{S}$ and $(d_{1}^{\prime},\dots,d_{n}^{\prime})=\overline{S^{\prime}}$. Let us prove:

[TABLE]

Let $L=\operatorname{{\rm lcm}}(S)$, $L^{\prime}=\operatorname{{\rm lcm}}(S^{\prime})$, and $L_{1}=\operatorname{{\rm lcm}}(S_{1})=\operatorname{{\rm lcm}}(S_{1}^{\prime})$. Let $g_{1}=g_{1}(S^{\prime})=g_{1}(S)$, i.e., $g_{1}=\gcd(a_{1}^{\prime},L_{1})=\gcd(a_{1},L_{1})$. We have $L=\operatorname{{\rm lcm}}(a_{1},L_{1})=a_{1}L_{1}/g_{1}$ and $L^{\prime}=\operatorname{{\rm lcm}}(a_{1}^{\prime},L_{1})=a_{1}^{\prime}L_{1}/g_{1}$, so for each $j\in\{1,\dots,n\}$, we have $d_{j}=L/a_{j}=\frac{a_{1}L_{1}}{a_{j}g_{1}}$ and $d_{j}^{\prime}=L^{\prime}/a_{j}^{\prime}=\frac{a_{1}^{\prime}L_{1}}{a_{j}^{\prime}g_{1}}$. This gives $d_{1}^{\prime}=L_{1}/g_{1}=d_{1}$ and, for $j\neq 1$, $d_{j}^{\prime}=\frac{a_{1}^{\prime}L_{1}}{a_{j}^{\prime}g_{1}}=\frac{ka_{1}L_{1}}{a_{j}g_{1}}=kd_{j}$; this proves the first part of (2). Since $\gcd(d_{1},k)$ is a divisor of $\gcd(d_{1},kd_{2},\dots,kd_{n})=\gcd(d_{1}^{\prime},\dots,d_{n}^{\prime})=1$, (2) is proved. Now let $\Phi:{\rm\bf k}[X_{1},\dots,X_{n}]\to{\rm\bf k}[Y_{1},\dots,Y_{n}]$ be the ${\rm\bf k}$-homomorphism that sends $X_{1}$ to $Y_{1}^{k}$ and $X_{j}$ to $Y_{j}$ for $j>1$. Noting that $Y_{1}^{a_{1}^{\prime}}+\cdots+Y_{n}^{a_{n}^{\prime}}=Y_{1}^{ka_{1}}+Y_{2}^{a_{2}}+\cdots+Y_{n}^{a_{n}}$, we see that \Phi^{-1}\big{(}\langle Y_{1}^{a_{1}^{\prime}}+\cdots+Y_{n}^{a_{n}^{\prime}}\rangle\big{)}=\langle X_{1}^{a_{1}}+\cdots+X_{n}^{a_{n}}\rangle, so $\Phi$ induces an injective homomorphism

[TABLE]

Then $B_{S}\cong\varphi(B_{S})={\rm\bf k}[y_{1}^{k},y_{2},\dots,y_{n}]$. Since $\deg(y_{j})=d_{j}^{\prime}$ for all $j$, (2) implies that $B_{S^{\prime}}^{(k)}={\rm\bf k}[y_{1}^{k},y_{2},\dots,y_{n}]$ and that $\varphi$ is homogeneous (meaning \varphi\big{(}(B_{S})_{i}\big{)}\subseteq(B_{S^{\prime}})_{ki} for all $i\in\mathbb{N}$); so $B_{S}\cong B_{S^{\prime}}^{(k)}$ as graded rings. Then $\operatorname{{\rm Proj}}B_{S}\cong\operatorname{{\rm Proj}}B_{S^{\prime}}$ follows from Proposition 5.1. ∎

5.3 Remark**.**

It is interesting to observe that the geometry of $\operatorname{{\rm Proj}}B_{S}$ is in general not sufficient to determine whether or not $B_{S}$ is rigid (whereas the geometry of $\operatorname{{\rm Spec}}B_{S}$ is of course sufficient). For example, let $S=(2,3,3,2)$, $S^{\prime}=(2,3,3,4)$ and $S^{\prime\prime}=(10,3,3,4)$. Since $S<^{4}S^{\prime}<^{1}S^{\prime\prime}$, Proposition 5.2 implies that $\operatorname{{\rm Proj}}B_{S}\cong\operatorname{{\rm Proj}}B_{S^{\prime}}\cong\operatorname{{\rm Proj}}B_{S^{\prime\prime}}$. However, $B_{S}=B_{2,3,3,2}$ is not rigid because $(2,3,3,2)\notin T_{4}$, and $B_{S^{\prime\prime}}=B_{10,3,3,4}$ is rigid by Theorem 1.3(e). Interestingly, we don’t know whether $B_{S^{\prime}}$ is rigid or not.

6. Stable Rigidity

6.1 Lemma**.**

Let $E\subseteq F$ be a field extension, $m\geq 1$, $t_{1},\dots,t_{m}$ independent indeterminates over $F$ and $f,g\in E[t_{1},\dots,t_{m}]\subseteq F[t_{1},\dots,t_{m}]$. Then $f,g$ are relatively prime in $E[t_{1},\dots,t_{m}]$ if and only if they are relatively prime in $F[t_{1},\dots,t_{m}]$.

Verification of the above Lemma is left to the reader. We quote Theorem 3.1 of [6]:

6.2 Theorem**.**

Let $m\geq 1$, $n\geq 3$, $g_{1},\dots,g_{n}\in\mathbb{C}[t_{1},\dots,t_{m}]=\mathbb{C}^{[m]}$ and $a_{1},\dots,a_{n}\in\mathbb{N}\setminus\{0\}$ be such that:

• •

$g_{1}^{a_{1}}+\cdots+g_{n}^{a_{n}}=0$**

• •

$g_{1},\dots,g_{n}$* are not all constant*

• •

given any $1\leq i_{1}<\dots<i_{s}\leq n$ such that $g_{i_{1}}^{a_{i_{1}}}+\cdots+g_{i_{s}}^{a_{i_{s}}}=0$, we have $\gcd(g_{i_{1}},\dots,g_{i_{s}})=1$.

Then $\sum_{i=1}^{n}\frac{1}{a_{i}}>\frac{1}{d-1}$ where $d$ is the dimension of the $\mathbb{C}$-vector space spanned by $g_{1}^{a_{1}},\dots,g_{n}^{a_{n}}$.

6.3 Corollary**.**

Let $K$ be a field of characteristic zero, $m\geq 1$, $n\geq 3$, $g_{1},\dots,g_{n}\in K[t_{1},\dots,t_{m}]=K^{[m]}$ and $a_{1},\dots,a_{n}\in\mathbb{N}\setminus\{0\}$ be such that:

• (i)

$g_{1}^{a_{1}}+\cdots+g_{n}^{a_{n}}=0$**

• (ii)

$\sum_{i=1}^{n}\frac{1}{a_{i}}\leq\frac{1}{n-2}$**

• (iii)

$g_{1},\dots,g_{n}$* are pairwise relatively prime.*

Then $g_{1},\dots,g_{n}\in K$.

Proof.

First consider the case where $K=\mathbb{C}$. Let $d$ be the dimension of the $\mathbb{C}$-vector space spanned by $g_{1}^{a_{1}},\dots,g_{n}^{a_{n}}$. Note that $d<n$ by (i); if $d\leq 1$ then the conclusion ($g_{1},\dots,g_{n}\in K$) immediately follows from (iii), so we may assume that $1<d<n$. Then $\frac{1}{d-1}$ is defined and $\frac{1}{d-1}\geq\frac{1}{n-2}$. So (ii) gives $\sum_{i=1}^{n}\frac{1}{a_{i}}\leq\frac{1}{d-1}$ and Theorem 6.2 implies that $g_{1},\dots,g_{n}\in K$. So the case $K=\mathbb{C}$ of Corollary 6.3 is true.

Now let $K$ be arbitrary. Let $K_{0}\subseteq K$ be the extension of $\mathbb{Q}$ generated by the coefficients of $g_{1},\dots,g_{n}$. Then $K_{0}$ can be embedded in $\mathbb{C}$; more precisely, if we choose a sufficiently large overfield $L$ of $K$ then we may find a copy of $\mathbb{C}$ in $L$ forming a diagram of fields as in the left part of:

[TABLE]

Now $g_{1},\dots,g_{n}\in K_{0}[t_{1},\dots,t_{m}]\subseteq\mathbb{C}[t_{1},\dots,t_{m}]$ are pairwise relatively prime in $\mathbb{C}[t_{1},\dots,t_{m}]$ by Lemma 6.1. So $g_{1},\dots,g_{n}$ satisfy (i–iii) as elements of $\mathbb{C}[t_{1},\dots,t_{m}]$. By the case $K=\mathbb{C}$ of Corollary 6.3, we get $g_{1},\dots,g_{n}\in\mathbb{C}$, so $g_{1},\dots,g_{n}\in K$. ∎

We need the notion of relatively prime elements of an arbitrary domain.

6.4 Definition**.**

Let $x,y\in B$ where $B$ is a domain. One says that $x,y$ are relatively prime in $B$ if the following hold:

• (i)

$xB\cap yB=xyB$   (ii) if $0\in\{x,y\}$ then $\{x,y\}\cap B^{*}\neq\varnothing$.

Note that this agrees with the usual notion when $B$ is a UFD.

6.5 Remark**.**

Let $x,y$ be relatively prime elements of a domain $B$.

1. (a)

If $S$ is a multiplicative set of $B$ such that $0\notin S$, then $x,y$ are relatively prime in $S^{-1}B$. 2. (b)

If $A$ is a factorially closed subring of $B$ such that $x,y\in A$, then $x,y$ are relatively prime in $A$. 3. (c)

If $B^{\prime}=B^{[N]}$ for some $N\geq 0$ then $x,y$ are relatively prime in $B^{\prime}$.

6.6 Definition**.**

Let $B$ be a domain of characteristic zero.

1. (a)

$\operatorname{{\rm ML}}(B)=\bigcap_{D\in\operatorname{{\rm LND}}(B)}\ker(D)$ 2. (b)

The rigid core of $B$ is defined as $\bigcap_{i=0}^{\infty}\operatorname{{\rm ML}}_{i}(B)$, where one defines $\operatorname{{\rm ML}}_{0}(B)=B$ and $\operatorname{{\rm ML}}_{i+1}(B)=\operatorname{{\rm ML}}(\operatorname{{\rm ML}}_{i}B)$ for all $i$.

The next result generalizes Theorem 6.1(a) of [10].

6.7 Theorem**.**

Let $B$ be a domain of characteristic zero, $n\geq 3$, $x_{1},\dots,x_{n}\in B$ and $a_{1},\dots,a_{n}\in\mathbb{N}\setminus\{0\}$. Assume that

• (i)

$x_{1}^{a_{1}}+\cdots+x_{n}^{a_{n}}=0$**

• (ii)

$\sum_{i=1}^{n}\frac{1}{a_{i}}\leq\frac{1}{n-2}$**

• (iii)

$x_{1},\dots,x_{n}$* are pairwise relatively prime in $B$.*

Then $x_{1},\dots,x_{n}\in\operatorname{{\rm ML}}(R)$ in each of the following cases:

1. (a)

$R$* is a factorially closed subring of $B$ satisfying $x_{1},\dots,x_{n}\in R$;* 2. (b)

$R=B^{[N]}$* for some $N$.*

Moreover, $x_{1},\dots,x_{n}$ belong to the rigid core of $B$.

Proof.

(a) Let $R$ be a factorially closed subring of $B$ satisfying $x_{1},\dots,x_{n}\in R$. Let $D\in\operatorname{{\rm LND}}(R)$ and let $A=\ker(D)$; we claim that $x_{1},\dots,x_{n}\in A$. To show this, we may assume that $D\neq 0$. Let $S=A\setminus\{0\}$, then $S^{-1}R=K^{[1]}$ where we set $K=\operatorname{{\rm Frac}}(A)$. Note that $x_{1},\dots,x_{n}$ are pairwise relatively prime in $R$ by Remark 6.5(b), so they are pairwise relatively prime in $S^{-1}R=K^{[1]}$ by Remark 6.5(a). Then $x_{1},\dots,x_{n}\in K$ by Corollary 6.3. As $A$ is factorially closed in $R$, we have $R\cap K=A$ and hence $x_{1},\dots,x_{n}\in A$. This argument shows that $x_{1},\dots,x_{n}\in\operatorname{{\rm ML}}(R)$, so case (a) is proved.

(b) Assume that $R=B^{[N]}$ for some $N$. Then $x_{1},\dots,x_{n}$ are pairwise relatively prime in $R$ by Remark 6.5(c). Applying part (a) to $x_{1},\dots,x_{n}\in R$ shows that $x_{1},\dots,x_{n}\in\operatorname{{\rm ML}}(R^{\prime})$ for any factorially closed subring $R^{\prime}$ of $R$ containing $x_{1},\dots,x_{n}$; in particular, $x_{1},\dots,x_{n}\in\operatorname{{\rm ML}}(R)$. So we are done in case (b).

Observe that $x_{1},\dots,x_{n}\in\operatorname{{\rm ML}}_{0}(B)$ and that if $i$ is such that $x_{1},\dots,x_{n}\in\operatorname{{\rm ML}}_{i}(B)$ then $x_{1},\dots,x_{n}\in\operatorname{{\rm ML}}_{i+1}(B)$ ($R=\operatorname{{\rm ML}}_{i}(B)$ is a factorially closed subring of $B$, so case (a) gives $x_{1},\dots,x_{n}\in\operatorname{{\rm ML}}(R)$). So $x_{1},\dots,x_{n}$ belong to the rigid core of $B$. ∎

One says that a ring $B$ of characteristic zero is stably rigid if $B\subseteq\operatorname{{\rm ML}}(R)$ for every overring $R$ of $B$ such that $R=B^{[N]}$ for some $N$. In the following statement, we set $B_{a_{1},\dots,a_{n}}={\rm\bf k}[X_{1},\dots,X_{n}]/\langle X_{1}^{a_{1}}+\cdots+X_{n}^{a_{n}}\rangle$ where ${\rm\bf k}$ is a field of characteristic zero.

6.8 Corollary**.**

Let $n\geq 3$ and $(a_{1},\dots,a_{n})\in(\mathbb{N}\setminus\{0\})^{n}$ be such that $\sum_{i=1}^{n}\frac{1}{a_{i}}\leq\frac{1}{n-2}$. Then $B_{a_{1},\dots,a_{n}}$ is stably rigid.

Proof.

The case $n=3$ is known (Theorem 7.1(b) of [10]), so we may assume that $n\geq 4$. Write $B=B_{a_{1},\dots,a_{n}}={\rm\bf k}[x_{1},\dots,x_{n}]$ where $x_{i}$ is the canonical image of $X_{i}$ in $B$. By Lemma 4.3, $x_{1},\dots,x_{n}$ are prime elements of $B$ and are pairwise relatively prime in $B$. Consider an overring $R=B^{[N]}$ of $B$ for some $N$. Then $x_{1},\dots,x_{n}\in\operatorname{{\rm ML}}(R)$ by case (b) of Theorem 6.7, so $B$ is stably rigid. ∎